Skip to main content
MathOverflow

Return to Answer

added 26 characters in body
Source Link
Serguei Popov
Serguei Popov
  • 1.9k
  • 12
  • 21

You can use the fact that $B_t-m_t$ is a reflected Brownian motion (see e.g. Revuz-Yor, Chapter VI, Theorem 2.3). I think it shouldn't be difficult to show that $$ \limsup \frac{M_t-m_t}{\sqrt{t\log\log t}} = \limsup \frac{B_t-m_t}{\sqrt{t\log\log t}}. $$$$ \limsup_{t\to\infty} \frac{M_t-m_t}{\sqrt{t\log\log t}} = \limsup_{t\to\infty} \frac{B_t-m_t}{\sqrt{t\log\log t}}. $$

You can use the fact that $B_t-m_t$ is a reflected Brownian motion (see e.g. Revuz-Yor, Chapter VI, Theorem 2.3). I think it shouldn't be difficult to show that $$ \limsup \frac{M_t-m_t}{\sqrt{t\log\log t}} = \limsup \frac{B_t-m_t}{\sqrt{t\log\log t}}. $$

You can use the fact that $B_t-m_t$ is a reflected Brownian motion (see e.g. Revuz-Yor, Chapter VI, Theorem 2.3). I think it shouldn't be difficult to show that $$ \limsup_{t\to\infty} \frac{M_t-m_t}{\sqrt{t\log\log t}} = \limsup_{t\to\infty} \frac{B_t-m_t}{\sqrt{t\log\log t}}. $$

Source Link
Serguei Popov
Serguei Popov
  • 1.9k
  • 12
  • 21

You can use the fact that $B_t-m_t$ is a reflected Brownian motion (see e.g. Revuz-Yor, Chapter VI, Theorem 2.3). I think it shouldn't be difficult to show that $$ \limsup \frac{M_t-m_t}{\sqrt{t\log\log t}} = \limsup \frac{B_t-m_t}{\sqrt{t\log\log t}}. $$